Bourbaki

Proposition 11

Tag 0AUR, permanent. Cite it as [Bourbaki, Tag 0AUR].

Lie Groups and Lie Algebras, chapter VI, § 1. Root systems

Assume that R is irreducible. Let α\alpha and β\beta be two roots such that ∥α∥=∥β∥\| \alpha \| = \| \beta \|. There exists g∈W(R)g \in W(R) such that g(α)=βg(\alpha) = \beta.

The transforms of α\alpha by W(R)W(R) generate V (no. 2, Cor. of Prop. 5). Hence there exists g∈W(R)g \in W(R) such that (g(α)∣β)≠0(g(\alpha)|\beta) \neq 0. We assume from now on that (α∣β)≠0(\alpha|\beta) \neq 0. By formula (9) of no. 1, n(α,β)=n(β,α)n(\alpha, \beta) = n(\beta, \alpha). Replacing β\beta if necessary by sβ(β)=−βs_\beta(\beta) = -\beta, we can assume that n(α,β)>0n(\alpha, \beta) > 0. Then, by the list at the beginning of no. 3, either α=β\alpha = \beta (in which case the proposition is clear), or n(α,β)=n(β,α)=1n(\alpha, \beta) = n(\beta, \alpha) = 1; in that case

sαsβsα(β)=sαsβ(β−α)=sα(−β−α+β)=α. s_\alpha s_\beta s_\alpha(\beta) = s_\alpha s_\beta(\beta - \alpha) = s_\alpha(-\beta - \alpha + \beta) = \alpha.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.